$AB=BA$ implies $AB^T=B^TA$ when $A$ is normal

I am looking for an elementary proof (if such exists) of the following: $$ AB=BA \quad\Longrightarrow\quad AB^T=B^TA, $$ where $A$ and $B$ are $n\times n$ real matrices, and $A$ is a normal matrix, i.e., $AA^T=A^TA$ - it is true for complex matrices as well, with $A^T$ replaced by $A^*$.

There is a non-elementary proof of this using the exponential of a matrix and properties of entire functions.

Update. In the first version of the question, both $A$ and $B$ were supposed to be normal, but as Shlomi correctly pointed out, this is true even in the case when only one of them is normal.

I'm not sure whether this qualifies as "elementary", but here's one way:

Note that $A$ and $B$ form a commuting family of matrices, so that $A$ and $B$ are simultaneously upper-triangularizable. That is, there is some unitary $U$ such that $U^*AU$ and $U^*BU$ are both upper triangular.

Note that $U^*AU$ and $U^*B U$ are both normal and upper-triangular, which means that they are diagonal. That is, we have $$ A = U\pmatrix{ \lambda_1 & &\\ &\ddots&\\ &&\lambda_n}U^* \quad B = U\pmatrix{ \mu_1 & &\\ &\ddots&\\ &&\mu_n}U^* $$ From here, it is easy to verify that $AB^* = B^*A$.

Alternatively, there's a nice inductive proof here if we use the following two facts:

• Commuting matrices have a common eigenvector
• A matrix of the form $\pmatrix{A&B\\0&C}$ is only normal if $B = 0$

With that, we may use the fact that $A$ and $B$ have a common eigenvector to write $$ A = U\pmatrix{ \lambda_1 & c^T\\ 0&\tilde A} U^* \quad B = U\pmatrix{ \mu_1 & d^T\\ 0&\tilde B}U^* $$ for a unitary $U$ (whose first column is a common eigenvector to $A$ and $B$). Because $A$ and $B$ are normal, we see that $c = d = 0$. Because $A$ and $B$ are commuting normal matrices, $\tilde A$ and $\tilde B$ are (smaller) commuting normal matrices.

By induction, we may conclude (as before) that for some unitary $U$, we have $$ A = U\pmatrix{ \lambda_1 & &\\ &\ddots&\\ &&\lambda_n}U^* \quad B = U\pmatrix{ \mu_1 & &\\ &\ddots&\\ &&\mu_n}U^* $$ and we may reach the desired conclusion, just like last time.

Proof when one of them is normal:

Assume that only $A$ is normal. So, there is an orthogonal matrix $P$ such that:

$A=P^*DP$, where $D$ is a diagonal matrix. We shall prove that there is a polynomial $q$, such that: $q(D)=D^*$. The fact, it is true from the interpolation polynomial. Hence, we conclude that: $q(A)=A^*$. Since $AB=BA$ we can prove that $q(A)B=Bq(A)$. Thus, $A^TB=BA^T$ and we are done.

It suffices to prove the complex case.

Since $A$ is normal, $Ax=\lambda x$ if and only if $A^\ast x=\overline{\lambda}x$.

Now, let $(\lambda,v)$ be an eigenpair of $A$. Then $ABv=BAv=B(\lambda v)=\lambda Bv$. By applying our previous argument to the cases $x=Bv$ and $x=v$ separately, we get $A^\ast Bv=\overline{\lambda}Bv$ and $A^\ast v=\overline{\lambda}v$. Combining the two, we obtain $A^\ast Bv=\overline{\lambda}Bv=B(\overline{\lambda}v)=BA^\ast v$.

Thus $A^\ast B$ and $BA^\ast$ agree at every eigenvector $v$ of $A$. Since $A$ is normal, it has a full eigenbasis. Therefore $A^\ast B=BA^\ast$ or equivalently, $AB^\ast=B^\ast A$.